| # | 제출 시각 | 아이디 | 문제 | 언어 | 결과 | 실행 시간 | 메모리 |
|---|---|---|---|---|---|---|---|
| 90034 | KLPP | 수열 (APIO14_sequence) | C++14 | 0 ms | 0 KiB |
이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#include<stdio.h>
#include<iostream>
#include<deque>
#include<queue>
using namespace std;
typedef long long int lld;
#define INF 1000000000000000000
typedef pair<lld,lld> line;
lld DP[100001][202];
int prev[100001][202];
lld sq(lld x){
return x*x;
}
class CH{
deque<pair<line,int> > d;
public:
void print(){
queue<pair<line,int> >q;
while(!d.empty()){
q.push(d.front());
cout<<d.front().first.first<<" "<<d.front().first.second<<" "<<d.front().second<<endl;
d.pop_front();
}
while(!q.empty()){
d.push_back(q.front());
q.pop();
}
}
double intersect(pair<line,int> a, pair<line,int> b){
return (double)(a.first.first-b.first.first)/(b.first.second-a.first.second);
}
void add(line l, int index){
pair<line,int> P;
P.first=l;
P.second=index;
while(d.size()>0 && P.first.second==d.back().first.second){
if(P.first.first<d.back().first.first){
d.pop_back();
}else{
return;
}
}
if(d.size()<2){
d.push_back(P);
return;
}
pair<line,int> A;
pair<line,int> B;
do{
A=d.back();
d.pop_back();
B=d.back();
//cout<<intersect(P,A)<<" "<<intersect(P,B)<<endl;
}while(d.size()>1 && intersect(P,A)<=intersect(P,B));
if(intersect(P,A)>intersect(P,B))d.push_back(A);
d.push_back(P);
}
int size(){
return d.size();
}
int query(lld c){
if(d.size()==1){
return d.front().second;
}
pair<line,int> A;
pair<line,int> B;
do{
A=d.front();
d.pop_front();
B=d.front();
}while(d.size()>1 && intersect(A,B)<=c);
if(intersect(A,B)>c)d.push_front(A);
return d.front().second;
}
void clear(){
while(!d.empty())d.pop_back();
}
};
int main(){
int n,k;
scanf("%d %d",&n,&k);
//cin>>n>>k;
lld arr[n];
lld acc[n+1];
for(int i=0;i<n;i++){
scanf("%lld",&arr[i]);
//cin>>arr[i];
}
acc[0]=0;
for(int i=0;i<n;i++){
acc[i+1]=acc[i]+arr[i];
}
for(int i=0;i<=n;i++){
for(int j=0;j<=k+1;j++){
DP[i][j]=INF;
}
}
/*CH *test=new CH();
test->add(line(1,-1),1);
test->add(line(1,-2),2);
test->add(line(3,-3),3);
test->print();*/
DP[0][0]=0;
CH *C=new CH();
for(int parts=0;parts<=k;parts++){
/*for(int i=0;i<n;i++){
for(int j=i+1;j<=n;j++){
if(DP[j][parts+1]>DP[i][parts]+sq(acc[j]-acc[i])){
DP[j][parts+1]=DP[i][parts]+sq(acc[j]-acc[i]);
prev[j][parts+1]=i;
}
}
}*/
for(int i=0;i<=n;i++){
if(C->size()>0){
/*cout<<i<<" "<<parts+1<<" "<<prev[i][parts+1]<<" "<<C->query(acc[i])<<endl;
if(prev[i][parts+1]!=C->query(acc[i])){
//C->print();
}*/
prev[i][parts+1]=C->query(acc[i]);
DP[i][parts+1]=DP[prev[i][parts+1]][parts]+sq(acc[i]-acc[prev[i][parts+1]]);
}
if(DP[i][parts]!=INF){
//cout<<i<<" "<<parts<<endl;
C->add(line(sq(acc[i])+DP[i][parts],-2*acc[i]),i);
//C->print();
//cout<<endl;
}
}//cout<<endl;
C->clear();
}
/*for(int parts=0;parts<=k+1;parts++){
for(int i=0;i<=n;i++){
printf("%d ",prev[i][parts]);
}printf("\n");
}*/
/*for(int i=1;i<=k+1;i++){
computeDP(0,n,i,0,n);
}*/
lld ans=sq(acc[n])-DP[n][k+1];
ans/=2;
printf("%lld\n",ans);
//cout<<ans<<endl;
int index=n;
int partition=k+1;
while(partition>0){
index=prev[index][partition];
partition--;
if(index>0){
printf("%d ",index);
//cout<<index<<" ";
}
}//cout<<endl;
printf("\n");
return 0;
}